Volume of Pyramid

Question 1

Find the volume of square based pyramid in the given figure alongside.

Solution:

Given information:

• Length of the side of base (a) = 16 cm
• Vertical height (h) = 6 cm
• Volume of pyramid (V) = ?

Step 1: Calculate the area of base

We know that area of base (A) = $a^2$

$\text{Area of base (A)} = a^2 = (16)^2 = 256 \text{ cm}^2$

Step 2: Calculate the volume of pyramid

We know that volume of pyramid (V) = $\frac{1}{3} \times A \times h$

Substituting the values:

$\text{Volume (V)} = \frac{1}{3} \times 256 \times 6$ $= \frac{1536}{3}$ $= 512 \text{ cm}^3$

Final Answer:

Hence, the volume of the pyramid is 512 cm³.

Verification:

• Base area: $a^2 = 16^2 = 256 \text{ cm}^2$ ✓
• Volume calculation: $V = \frac{1}{3} \times 256 \times 6 = \frac{1536}{3} = 512 \text{ cm}^3$ ✓
• Unit check: Area (cm²) × Height (cm) ÷ 3 = Volume (cm³) ✓

Volume formula for square-based pyramid:

• General formula: $V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$
• For square base: $V = \frac{1}{3} \times a^2 \times h$ where $a$ is base side and $h$ is vertical height
• Alternative form: $V = \frac{a^2 h}{3}$

Key concepts:

• Pyramid volume: Always one-third of the corresponding prism volume
• Base area: For square base, simply square the side length
• Vertical height: The perpendicular distance from apex to base center
• Units: Volume is always in cubic units (length³)

Note: The volume of any pyramid is exactly one-third the volume of a prism with the same base and height. This is a fundamental relationship in solid geometry that applies to pyramids of any base shape.

Question 2

The volume of a square based pyramid is 384 cm³ and the length of the side of base is 12 cm. Find the lateral surface area of the pyramid.

Solution:

Given information:

• Volume of pyramid (V) = 384 cm³
• Length of the side of base (a) = 12 cm
• Lateral surface area = ?

Step 1: Find the vertical height (h)

We know that volume of the pyramid (V) = $\frac{1}{3} \times a^2 \times h$

Substituting the given values:

$384 = \frac{1}{3} \times (12)^2 \times h$ $384 = \frac{1}{3} \times 144 \times h$ $384 = \frac{144h}{3}$ $\frac{384 \times 3}{144} = h$ $h = 8$ cm

∴ The vertical height (h) = 8 cm

Step 2: Find the slant height (l)

Using the relationship: $l^2 = h^2 + \left(\frac{a}{2}\right)^2$

Substituting the values:

$l^2 = 8^2 + \left(\frac{12}{2}\right)^2$ $l^2 = 64 + 36$ $l^2 = 100$ $l = 10$ cm

∴ The slant height (l) = 10 cm

Step 3: Calculate lateral surface area

Lateral surface area (LSA) = $2al = 2 \times 12 \times 10 = 240$ cm²

Final Answer:

Hence, the lateral surface area of the pyramid is 240 cm².

Verification:

• Volume check: $V = \frac{1}{3} \times 12^2 \times 8 = \frac{1152}{3} = 384 \text{ cm}^3$ ✓
• Slant height check: $l^2 = h^2 + \left(\frac{a}{2}\right)^2 = 8^2 + 6^2 = 64 + 36 = 100$ ∴ $l = 10$ cm ✓
• Lateral surface area: $LSA = 2al = 2 \times 12 \times 10 = 240 \text{ cm}^2$ ✓
• Pythagorean triple: 6-8-10 triangle (scaled 3-4-5) ✓

Problem-solving sequence:

• Step 1: Use volume formula to find vertical height from given volume and base
• Step 2: Apply Pythagorean theorem to find slant height from vertical height
• Step 3: Calculate lateral surface area using slant height and base perimeter

Key relationships used:

• Volume formula: $V = \frac{1}{3} \times a^2 \times h$ (base area × height ÷ 3)
• Pythagorean relationship: $l^2 = h^2 + \left(\frac{a}{2}\right)^2$ (slant height, vertical height, half-base)
• Lateral surface area: $\text{LSA} = 2al$ (perimeter × slant height ÷ 2)

Note: This is a reverse problem where we work backwards from volume to find height, then use geometric relationships to find slant height and finally lateral surface area. The 6-8-10 right triangle formed is a common Pythagorean triple that provides easy verification.

Question 3

Find the volume of a square based pyramid if its total surface area is 96 cm² and the length of the side of base is 6 cm.

Solution:

Given information:

• Total surface area (TSA) = 96 cm²
• Length of the side of base (a) = 6 cm
• Volume of pyramid (V) = ?

Step 1: Find the slant height (l)

According to the formula, total surface area (TSA) = $a^2 + 2al$

Substituting the given values:

$96 = 6^2 + 2 \times 6 \times l$ $96 = 36 + 12l$ $96 - 36 = 12l$ $60 = 12l$ $l = \frac{60}{12} = 5$ cm

∴ The slant height (l) = 5 cm

Step 2: Find the vertical height (h)

Using the relationship: $l^2 = h^2 + \left(\frac{a}{2}\right)^2$

Substituting the values:

$5^2 = h^2 + \left(\frac{6}{2}\right)^2$ $25 = h^2 + 9$ $25 - 9 = h^2$ $h^2 = 16$ $h = 4$ cm

∴ The vertical height (h) = 4 cm

Step 3: Calculate the volume of pyramid

Volume of pyramid (V) = $\frac{1}{3} \times \text{area of base} \times \text{height}$

Substituting the values:

$V = \frac{1}{3} \times 6^2 \times 4$ $= \frac{1}{3} \times 36 \times 4$ $= \frac{144}{3}$ $= 48$ cm³

Final Answer:

Hence, the volume of the pyramid is 48 cm³.

Verification:

• Total surface area check: $TSA = a^2 + 2al = 6^2 + 2 \times 6 \times 5 = 36 + 60 = 96 \text{ cm}^2$ ✓
• Pythagorean relationship: $l^2 = h^2 + \left(\frac{a}{2}\right)^2 = 4^2 + 3^2 = 16 + 9 = 25 = 5^2$ ✓
• Volume calculation: $V = \frac{1}{3} \times 36 \times 4 = \frac{144}{3} = 48 \text{ cm}^3$ ✓
• Pythagorean triple: 3-4-5 triangle (classic Pythagorean triple) ✓

Problem-solving approach:

• Step 1: Use total surface area formula to find slant height from given TSA and base
• Step 2: Apply Pythagorean theorem to find vertical height from slant height
• Step 3: Calculate volume using base area and vertical height

Key formulas used:

• Total surface area: $\text{TSA} = a^2 + 2al$ (base area + lateral surface area)
• Pythagorean relationship: $l^2 = h^2 + \left(\frac{a}{2}\right)^2$ (right triangle formed)
• Volume formula: $V = \frac{1}{3} \times a^2 \times h$ (pyramid volume)

Geometric insight:

• Surface area constraint: Given TSA provides direct path to slant height
• Right triangle formation: Vertical height, half-base (3 cm), and slant height (5 cm) form 3-4-5 triangle
• Classic Pythagorean triple: The 3-4-5 triangle is the most common right triangle in geometry

Note: This problem demonstrates the interconnected nature of pyramid measurements. Starting with total surface area, we can systematically find slant height, then vertical height, and finally volume. The appearance of the 3-4-5 Pythagorean triple makes verification straightforward.